Question

Chứng tỏ rằng :

(1+\(\dfrac{1}{3}\)+\(\dfrac{1}{5}\)+...+\(\dfrac{1}{99}\)) - (\(\dfrac{1}{2}\)+\(\dfrac{1}{4}\)+\(\dfrac{1}{6}\)+...+\(\dfrac{1}{100}\)) = \(\dfrac{1}{51}\)+ \(\dfrac{1}{52}\)+ \(\dfrac{1}{53}\)+ ...+ \(\dfrac{1}{100}\)

Answer 1

\(\left(1+\dfrac{1}{3}+\dfrac{1}{5}+.....+\dfrac{1}{99}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+.......+\dfrac{1}{100}\right)\)

\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+.....+\dfrac{1}{99}+\dfrac{1}{100}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+.......+\dfrac{1}{100}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+.....+\dfrac{1}{100}\right)\)

\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+.....+\dfrac{1}{100}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+.......+\dfrac{1}{100}\right)\)

\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{100}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+.....+\dfrac{1}{50}\right)\)

\(=\dfrac{1}{51}+\dfrac{1}{52}+......+\dfrac{1}{100}\)